Logarithms of readings of pressure and volume for an ideal gas were plotted on a graph as shown in Fig. By measuring the gradient, it can be shwon that the gas may be

Options

A. Monoatomic and undergoing an adiabatic change.
B. Monoatomic and undergoing an isothermal change.
C. Diatomic and undergoing an adiabatic change.
D. Triatomic and undergoing an isothermal change.

Answer

C

Explanation

`PV^(gamma) = C, ln P+ gammalnV = lnC`
` implies lnP = -gammalnV + lnC implies y = mx +c`
` implies m = -gamma = -(-[2.10 - 2.38])/((1.30 - 1.10)) = -1.4`
`therefore` The gas is diatomic